Question

Suppose two people (let’s call them Julio and Karina) agree to meet for lunch at a certain restaurant, each person’s arrival time, in minutes after noon, follows a normal distribution with mean 30 and standard deviation 10. Assume that they arrive independently of each other and that they agree to wait for 15 minutes. If each person agrees to wait exactly fifteen minutes for the other before giving up and leaving.

h) Report the probability distribution of the difference (not absolute difference) in the arrival times of Julio and Karina. [Hint: You might let Tj represent Julio’s arrival time and Tk represent Karina’s arrival time, both in minutes after noon. Use what you know about normal distributions to specify the probability distribution of the difference D = Tj – Tk.]

i) Use appropriate normal probability calculations to determine the probability that the two people successfully meet. Also report the values of the appropriate z-scores. [Hint: First express the probability that they successfully meet in terms of the random variable D.]

j) Now let m represent the number of minutes that both people agree to wait, where m can be any real number. Determine the value of m so the probability of meeting is .9

k) Now suppose that Julio and Karina can only afford to wait for 15 minutes, but they want to have at least a 90% chance of successfully meeting. Continue to assume that their arrival times follow independent normal distributions with mean 30 and the same SD as each other. Determine how small that SD needs to be in order to meet their criteria. (As always, show your work.)

Answer 1

h)

Tj ~ N(30, 10²) and  Tk ~ N(30, 10²)

D = Tj – Tk

We know that the linear combination of Normal random variable is a normal random variable.

E(Tj – Tk) = E(Tj) – E(Tk) = 30 - 30 = 0

Var(Tj – Tk) = Var(Tj) + E(Tk) = 10² + 10² = 200 (Both arrivals are independent of each other)

Thus,  D = Tj – Tk ~ N(0, 200)

i)

Probability that the two people successfully meet = P(-15 < D < 15)

Z score for D = -15 is (-15 - 0) / = -1.06

Z score for D = 15 is (15 - 0) / = 1.06

Probability that the two people successfully meet = P(-1.06 < Z < 1.06)

= P(Z < 1.06) - P(Z < -1.06)

= 0.8554 - 0.1446

= 0.7108

j)

Probability that the two people successfully meet = P(-m < D < m) = 0.9

Using symmetry of Normal distribution,

P(D > m) = (1 - 0.9)/2 = 0.05

Z score for D = m is (m - 0) / = m/14.142

P(Z > m/14.142) = 0.05

=> m/14.142 = 1.645

=> m = 14.142 * 1.645 = 23.26 minutes

k)

Probability that the two people successfully meet = P(-15 < D < 15) = 0.9

Using symmetry of Normal distribution,

P(D > 15) = (1 - 0.9)/2 = 0.05

Z score for D = 15 is (15 - 0) / S = 15/S where S is the standard deviation of D

P(Z > 15/S) = 0.05

=> 15/S = 1.645

=> S = 15/1.645 = 9.12 minutes

Now, S = = SD

=> SD = S/ = 9.12 / = 6.45 minutes